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Question 24

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Find the equation of the locus of a point P(x, y), which is equidistant from Q (0, 0) and R (2, 1)

Option A:

4x +2y = 5

Option B:

2x + 2y = 5

Option C:

4x – 2y = 5

Option D:

2x + y = 5

Jamb Maths Solution:

Note: The locus of P(x,y) which is equidistant from Q and R is the perpendicular bisector of the line QRMeaning we have to find the mid-point of the line line QR, obtain the slope (or gradient) of line QR from which can use to determine the slope CP from condition for two lines to be perpendicular (${{m}_{1}}{{m}_{2}}=-1$). Then use one – point slope equation of a line to determine the equation of the line.Follow the workings below$\begin{align} & \text{Mid}-\text{point of }QR \\ & (x,y)=\left( \frac{0+2}{2},\frac{0+1}{2} \right)=\left( 1,\tfrac{1}{2} \right) \\ & \text{Slope of line }QR \\ & {{m}_{1}}=\frac{1-0}{2-0}=\frac{1}{2} \\ & \text{The line }QR\text{ must be }\bot \text{ar to }CP \\ & \text{For two line to }\bot \text{ar, }{{m}_{1}}{{m}_{2}}=-1 \\ & \text{Let the slope of }CD={{m}_{2}} \\ & \frac{1}{2}\times {{m}_{2}}=-1 \\ & {{m}_{2}}=-2 \\ & \text{Equation of line }CD\text{ passing through }\left( 1,\tfrac{1}{2} \right)\text{ using } \\ & \text{one point slope form equation} \\ & y-{{y}_{1}}=m(x-{{x}_{1}}) \\ & y-\tfrac{1}{2}=-2(x-1) \\ & y-\tfrac{1}{2}=-2x+2 \\ & 2y-1=-4x+4 \\ & 4x+2y=5 \\\end{align}$